Possible mistake during differentiation? Please check

  • Context: Undergrad 
  • Thread starter Thread starter Dominathan
  • Start date Start date
  • Tags Tags
    Differentiation Mistake
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
Dominathan
Messages
2
Reaction score
0
For a function f(x), I have to determine intervals of increase/decrease, find local max(s)/min(s), and find intervals of concavity. The first thing I'm doing in this is to write out f'(x) and f''(x).

f(x) = [itex]ln(x)/\sqrt{x}[/itex]

For f'(x), I used the quotient rule and received f'(x) = (([itex]\frac{1}{x}[/itex][itex]\sqrt{x}[/itex])-([itex]\frac{-\sqrt{x}}{2}[/itex]ln(x))) / 2

However, I plugged f(x) into wolfram alpha and it gave me: [itex]\frac{2-ln(x)}{2x^{3/2}}[/itex]

I don't understand the difference? I thought I had done this correctly but apparently not? Wolfram alpha used the product rule. Is there some kind of algebraic gymnastics I'm forgetting about? I really want to understand where my error was made, not just which is the correct answer. Thanks!
 
Physics news on Phys.org
In using the quotient rule you are not differentiating [itex]\sqrt{x}[/itex] corretly and the quotient rule requires you to divide by the square of the denominator.
 
Dominathan said:
f(x) = [itex]ln(x)/\sqrt{x}[/itex]

For f'(x), I used the quotient rule and received f'(x) = (([itex]\frac{1}{x}[/itex][itex]\sqrt{x}[/itex])-([itex]\frac{-\sqrt{x}}{2}[/itex]ln(x))) / 2

I see a couple of issues here:

1) How did you get [itex]\frac{-\sqrt{x}}{2}[/itex] in the numerator? What is the derivative of [itex]\sqrt{x}[/itex] ?

2) How did you get 2 in the denominator? [itex](\sqrt{x})^{2} =[/itex] ?
 
Using quotient rule, you should get [itex]\frac{\frac{\sqrt{x}}{x} - \frac{\ln{x}}{2 \sqrt{x}}}{x}[/itex] which simplifies to what you got from WA. It looks like you messed up on the derivative of [itex]\sqrt{x}[/itex] and on the bottom of the quotient rule. http://en.wikipedia.org/wiki/Quotient_rule
 
gb7nash said:
I see a couple of issues here:

1) How did you get [itex]\frac{-\sqrt{x}}{2}[/itex] in the numerator? What is the derivative of [itex]\sqrt{x}[/itex] ?2) How did you get 2 in the denominator? [itex](\sqrt{x})^{2} =[/itex] ?

1.
The following is my logic for the answer I received:
  1. [itex]\sqrt{x}[/itex] = x[itex]^{1/2}[/itex]
  2. Using the power rule I bring the 1/2 out as a coefficient, and subtract one from the numerator : [itex]\frac{1}{2}[/itex][itex]x^{-1/2}[/itex]
  3. I simplified to : [itex]\frac{-\sqrt{x}}{2}[/itex]

2.
My bad! I did a poor job transcribing it from my notebook to the syntax used on this site. It was (obviously) my first post, but far from my last! I meant to put "x" as the denominator, that was a mental slip.
 
Welcome to the PhysicsForums! One may sometimes swallow some water but that is the way to learns to swim!